HOC in the example is not pure function with zero side effects?

To me the statement on https://github.com/reactjs/reactjs.org/blob/master/content/docs/higher-order-components.md appear to be untrue:
A HOC is a pure function with zero side-effects.
where just above the HOC is returning container that surely has side effects: mutation of state:

// This function takes a component...
function withSubscription(WrappedComponent, selectData) {
  // ...and returns another component...
  return class extends React.Component {
    constructor(props) {
      super(props);
      this.handleChange = this.handleChange.bind(this);
      this.state = {
        data: selectData(DataSource, props)
      };
    }

    componentDidMount() {
      // ... that takes care of the subscription...
      DataSource.addChangeListener(this.handleChange);
    }

    componentWillUnmount() {
      DataSource.removeChangeListener(this.handleChange);
    }

    handleChange() {
      this.setState({
        data: selectData(DataSource, this.props)
      });
    }

    render() {
      // ... and renders the wrapped component with the fresh data!
      // Notice that we pass through any additional props
      return <WrappedComponent data={this.state.data} {...this.props} />;
    }
  };
}

Do you agree?

从 content/docs/higher-order-components.md 中“一个 HOC 是没有任何副作用的纯函数”这一表述开始，并将其与其中包含的 withSubscription 示例进行比较。查看周围的说明和包含五条评论的讨论，然后明确措辞，使该表述与示例保持一致。当文档准确描述示例所展示的副作用时，即视为完成。